What Is Long Division of Polynomials?
Long division of polynomials divides a polynomial, called the dividend, by a polynomial of equal or smaller degree, called the divisor, to get a quotient and a remainder. When the divisor has more than one term, long division is the standard tool, where shortcuts like synthetic division no longer apply.
The relationship it builds is the division algorithm: for a dividend $P(x)$ and divisor $D(x)$,
$$P(x) = D(x) \cdot Q(x) + R(x)$$
where $Q(x)$ is the quotient and $R(x)$ is the remainder, whose degree is always less than the degree of $D(x)$. That last condition, remainder degree below divisor degree, is the signal that tells you when to stop.
Variable Glossary:
Symbol | Meaning |
|---|---|
$P(x)$ | dividend (the polynomial being divided) |
$D(x)$ | divisor (what you divide by) |
$Q(x)$ | quotient (the answer on top) |
$R(x)$ | remainder (what's left, lower degree than $D$) |
How Do You Do Long Division of Polynomials Step by Step?
The method is five repeating moves. Run them until the remainder's degree falls below the divisor's.
Arrange both polynomials in standard form, highest power to lowest. Insert a zero coefficient for any missing power.
Divide the leading term of the dividend by the leading term of the divisor. Write the result in the quotient.
Multiply the whole divisor by that quotient term.
Subtract that product from the dividend.
Bring down the next term and repeat from step 2.
Watch it on $(x^2 + 5x + 6) \div (x + 2)$.
$$x^2 \div x = x$$
$$x \cdot (x + 2) = x^2 + 2x$$
$$(x^2 + 5x) - (x^2 + 2x) = 3x$$
Bring down the $+6$ to get $3x + 6$.
$$3x \div x = 3$$
$$3 \cdot (x + 2) = 3x + 6$$
$$(3x + 6) - (3x + 6) = 0$$
The quotient is $x + 3$, remainder $0$.
What Do You Do with Missing Terms?
A missing power is the single most common reason a long division goes sideways. If the dividend jumps from $x^4$ straight to $x^2$, the $x^3$ term is missing, and you must hold its place with a zero coefficient before dividing.
Rewrite $6x^4 - 9x^2 + 3x + 6$ as $6x^4 + 0x^3 - 9x^2 + 3x + 6$. The $0x^3$ adds nothing to the value but keeps your columns aligned, the same way a zero holds the tens place in the number $304$. Skip it, and a later subtraction lands in the wrong column and the whole answer drifts.
Dividing by a Monomial Versus a Binomial
Not every division needs the full bracket. When the divisor is a single term, you can split the work; when it has two or more terms, the long-division layout earns its place.
Divisor is a monomial (one term, like $2x$): divide each term of the dividend separately. $(4x^2 - 6x) \div 2x = \frac{4x^2}{2x} - \frac{6x}{2x} = 2x - 3$.
Divisor is a binomial or larger (like $x - 3$ or $x^2 + 1$): use the full long-division algorithm, because the terms interact and cannot be split apart.
This connects to factoring: if you know one root $r$ of a polynomial, dividing by $(x - r)$ peels off a factor and drops the degree by one, which is how factoring cubic polynomials often begins.
Examples of Long Division of Polynomials
Six examples, from a clean exact division to a missing-term divide with a remainder. The problem is bold; the steps are not.
Example 1
Divide $(2x^2 + 3x - 2) \div (x + 2)$.
$$2x^2 \div x = 2x$$
$$2x \cdot (x + 2) = 2x^2 + 4x$$
$$(2x^2 + 3x) - (2x^2 + 4x) = -x$$
Bring down $-2$: now $-x - 2$.
$$-x \div x = -1$$
$$-1 \cdot (x + 2) = -x - 2$$
$$(-x - 2) - (-x - 2) = 0$$
Quotient $2x - 1$, remainder $0$.
Example 2
Divide $(4x^2 - 5x - 21) \div (x - 3)$, after first seeing a tempting wrong move.
Wrong attempt: a student divides $4x^2$ by $x$ to get $4x$, multiplies to get $4x^2 - 12x$, then subtracts only the leading term and writes the next line as $-5x$ instead of $-5x - (-12x)$.
That drops the $+12x$ that the subtraction creates, and the working collapses to a wrong quotient. The flaw is subtracting one term instead of the whole product.
Correct method: subtract the entire product each time.
$$4x^2 \div x = 4x$$
$$4x \cdot (x - 3) = 4x^2 - 12x$$
$$(4x^2 - 5x) - (4x^2 - 12x) = 7x$$
Bring down $-21$: now $7x - 21$.
$$7x \div x = 7$$
$$7 \cdot (x - 3) = 7x - 21$$
$$(7x - 21) - (7x - 21) = 0$$
Quotient $4x + 7$, remainder $0$.
Example 3
Divide $(x^3 + 0x^2 + 0x - 8) \div (x - 2)$, a perfect-cube-style dividend with missing terms.
Placeholders are already inserted for the missing $x^2$ and $x$ terms.
$$x^3 \div x = x^2$$
$$x^2 \cdot (x - 2) = x^3 - 2x^2$$
$$(x^3 + 0x^2) - (x^3 - 2x^2) = 2x^2$$
Bring down $0x$: now $2x^2 + 0x$.
$$2x^2 \div x = 2x$$
$$2x \cdot (x - 2) = 2x^2 - 4x$$
$$(2x^2 + 0x) - (2x^2 - 4x) = 4x$$
Bring down $-8$: now $4x - 8$.
$$4x \div x = 4$$
$$4 \cdot (x - 2) = 4x - 8$$
$$(4x - 8) - (4x - 8) = 0$$
Quotient $x^2 + 2x + 4$, remainder $0$.
Example 4
Divide $(2x^2 + 5x + 1) \div (x + 1)$ when the division does not come out even.
$$2x^2 \div x = 2x$$
$$2x \cdot (x + 1) = 2x^2 + 2x$$
$$(2x^2 + 5x) - (2x^2 + 2x) = 3x$$
Bring down $+1$: now $3x + 1$.
$$3x \div x = 3$$
$$3 \cdot (x + 1) = 3x + 3$$
$$(3x + 1) - (3x + 3) = -2$$
The remainder $-2$ has degree $0$, below the divisor's degree $1$, so stop.
Answer: $2x + 3 - \dfrac{2}{x + 1}$.
Example 5
Divide a polynomial by a quadratic: $(6x^3 + 12x^2 + 2x + 25) \div (x^2 + 4x + 3)$.
$$6x^3 \div x^2 = 6x$$
$$6x \cdot (x^2 + 4x + 3) = 6x^3 + 24x^2 + 18x$$
$$(6x^3 + 12x^2 + 2x) - (6x^3 + 24x^2 + 18x) = -12x^2 - 16x$$
Bring down $+25$: now $-12x^2 - 16x + 25$.
$$-12x^2 \div x^2 = -12$$
$$-12 \cdot (x^2 + 4x + 3) = -12x^2 - 48x - 36$$
$$(-12x^2 - 16x + 25) - (-12x^2 - 48x - 36) = 32x + 61$$
The remainder $32x + 61$ has degree $1$, below the divisor's degree $2$, so stop.
Answer: $6x - 12 + \dfrac{32x + 61}{x^2 + 4x + 3}$.
Example 6
Divide $(6x^4 - 9x^2 + 3x + 6) \div (x^2 - 2)$, a missing-term case.
Insert the placeholder: $6x^4 + 0x^3 - 9x^2 + 3x + 6$.
$$6x^4 \div x^2 = 6x^2$$
$$6x^2 \cdot (x^2 - 2) = 6x^4 - 12x^2$$
$$(6x^4 + 0x^3 - 9x^2) - (6x^4 - 12x^2) = 0x^3 + 3x^2$$
Bring down $3x$: now $3x^2 + 3x$.
$$3x^2 \div x^2 = 3$$
$$3 \cdot (x^2 - 2) = 3x^2 - 6$$
$$(3x^2 + 3x + 6) - (3x^2 - 6) = 3x + 12$$
The remainder $3x + 12$ has degree $1$, below the divisor's degree $2$, so stop.
Answer: $6x^2 + 3 + \dfrac{3x + 12}{x^2 - 2}$.
How to Check Your Answer
Every long division can be checked, and the check is the division algorithm read backwards: multiply the quotient by the divisor, add the remainder, and you must get the original dividend back. If you do not, the division has an error somewhere.
The relationship to verify is:
$$D(x) \cdot Q(x) + R(x) = P(x)$$
Take Example 4, where $(2x^2 + 5x + 1) \div (x + 1)$ gave quotient $2x + 3$ and remainder $-2$.
Multiply quotient by divisor:
$$(2x + 3)(x + 1) = 2x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3$$
Add the remainder:
$$2x^2 + 5x + 3 + (-2) = 2x^2 + 5x + 1$$
That is the original dividend, so the answer checks. Running this multiply-back step takes under a minute and catches the dropped-sign and missing-term errors before they cost marks. Make it a habit on every division, not just the ones you are unsure about.
Why Long Division of Polynomials Exists
Long division was not invented to make algebra homework longer. It exists because dividing polynomials is the only reliable way to break a complicated expression into simpler factors when no obvious shortcut is available.
It finds factors from known roots. If $r$ is a root of $P(x)$, then $(x - r)$ divides it cleanly, and long division produces the smaller polynomial $Q(x)$ that hides the remaining roots. This is the engine behind solving higher-degree equations.
It powers the remainder theorem. The remainder when you divide $P(x)$ by $(x - r)$ equals $P(r)$, a fact that turns a long division into a quick root check.
It simplifies rational functions. Before integrating or graphing a ratio of polynomials, engineers and analysts long-divide to separate the polynomial part from the proper fraction left over, which is exactly the form Example 5 produced.
The deeper reason is that polynomials behave like whole numbers under division: there is always a unique quotient and remainder. Long division is the procedure that makes that guarantee usable by hand.
Where Solutions Go Off the Rails
Most long-division errors are bookkeeping errors, not algebra errors. Three patterns catch students again and again.
Mistake 1: Subtracting only the leading term
Where it slips in: After multiplying the divisor by the quotient term, at the subtraction step.
Don't do this: Subtract just the first term of the product and copy the rest down unchanged.
The correct way: Subtract the entire product, every term, then bring down the next. The first instinct is to subtract one term and forget that the subtraction changes the sign of every term in the product, and that dropped sign change is where the answer breaks.
Mistake 2: Skipping the zero placeholder for a missing term
Where it slips in: When the dividend jumps a power, like going from $x^4$ to $x^2$.
Don't do this: Write the dividend with the gap left open and divide straight into it.
The correct way: Insert $0x^3$ (or whatever power is missing) to hold the column. The rusher who skips this saves three seconds and then loses ten minutes when the subtractions land in the wrong columns.
Mistake 3: Stopping too early or too late
Where it slips in: Deciding when the division is finished.
Don't do this: Stop the moment the numbers look small, or keep dividing past a valid remainder.
The correct way: Stop only when the remainder's degree is strictly less than the divisor's degree. A remainder of $3x + 12$ is final against a divisor of degree $2$, but not against a divisor of degree $1$.
Key Takeaways
Long division of polynomials repeats divide, multiply, subtract, bring down until the remainder's degree drops below the divisor's.
Always arrange both polynomials in standard form and insert zero placeholders for missing terms.
The result satisfies the division algorithm $P(x) = D(x) Q(x) + R(x)$, which you can use to check your answer.
Subtract the whole product each cycle, not just its leading term.
Divide by a monomial term by term; reserve full long division for binomial and larger divisors.
Practice This to Solidify Your Understanding
Work these three: (1) $(x^2 + 7x + 10) \div (x + 2)$, (2) $(2x^3 - 3x + 5) \div (x - 1)$ with placeholders inserted, (3) $(x^3 - 1) \div (x^2 + x + 1)$. Set each up in standard form first, insert zeros for any missing power, and stop only when the remainder's degree drops below the divisor's. If a subtraction confuses you, return to the step-by-step section above. To go faster on $x - k$ divisors, look next at dividing polynomials and synthetic division.
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